Class 11 Nootan Physics Solution

Explanation: This asks which factors determine the number of Electric Field lines from a point charge. Electric Field lines depict the direction and strength of the Electric Field. The total flux from a charge is proportional to the magnitude of the charge and influenced by the surrounding dielectric medium. According to Gauss’s law, flux Φ = Q/ε, where Q is the enclosed charge and ε is the permittivity of the medium. A larger charge produces more field lines, while the dielectric affects their distribution, not the total number. The surface area around the charge affects flux density but not the number of lines themselves. Analogy: A bright bulb emits more rays in all directions than a dim bulb, just as a larger charge emits more field lines. In summary, the number of tubes depends mainly on the charge, with medium affecting spacing.

Explanation: This examines how inserting a dielectric changes voltage across a charged Capacitor. Capacitance increases as C = K × C₀, where K is the dielectric constant. For a disconnected Capacitor, charge Q remains constant. Initially, V = 100 V with capacitance C₀. With dielectric, C = 10 × C₀, so new voltage V’ = Q / (10 × C₀), reducing voltage by a factor equal to K. Analogy: Inflating a balloon (charge) that now stretches more easily (dielectric) lowers internal pressure (voltage). Summary: Inserting a dielectric reduces voltage while increasing capacitance proportionally to K.

Explanation: This problem involves combining Capacitors in parallel. When disconnected from the battery, the first Capacitor retains its charge Q = C₁V₁. Connecting a second Capacitor C₂ in parallel redistributes the total charge: Q = (C₁ + C₂)V_f. Given V_f = 30 V, we can relate C₂ = (C₁V₁ / V_f) – C₁. Key concept: charge conservation. Analogy: Water in one tank flows into a second tank until both levels are equal, like charge distributing across parallel Capacitors. Summary: Using Q = C V relationships allows calculation of the unknown capacitance.

Explanation: When a capacitor is connected to a battery, it draws charge until it reaches the battery voltage. energy supplied by the battery is partially stored as electric potential energy U = ½ CV² in the capacitor; the remainder is dissipated as Heat in the connecting wires. Key point: only half of the supplied energy is stored; the rest is lost in the charging process. Analogy: Pumping water into a tank with resistance in the pipe wastes some energy as Heat. Summary: Stored energy depends on capacitance and voltage, while energy supplied is higher due to losses.

Explanation: This problem requires analyzing capacitor interaction. When connected, the total charge redistributes. For both potentials to drop to zero, the Capacitors must have charges of equal magnitude but opposite sign initially. Using Q = CV, the relation C₁V₁ = C₂V₂ provides the proportion of capacitances. Analogy: Two water tanks at different pressures are connected; equilibrium occurs when pressures equalize. Summary: Charge conservation and Q = CV allow determination of the required capacitance ratio.

Explanation: This question uses electric flux concepts. Total flux through a sphere is Φ = Q/ε₀. A portion subtending Ω steradians has flux Φ_p = (Ω / 4π) × Q/ε₀. With Ω = π/2, flux is (1/8) of total. Key idea: flux is proportional to the Solid angle subtended. Analogy: A pie chart slice represents a fraction of total pie, like a portion of flux through a sphere. Summary: Electric flux through a portion depends on its Solid angle relative to the full sphere.

Explanation: When the charged capacitor is shorted, the stored energy U = ½ CV² is released as Heat in the wire. This comes from the energy initially stored in the Electric Field between the plates. Key point: All energy stored in the capacitor converts to Heat, not mechanical work. Analogy: A stretched spring releases all potential energy as Heat if it hits a frictional surface. Summary: Heat generated equals the initial energy stored in the capacitor.

Explanation: work done in creating a bubble is proportional to its surface area, W = 2 × surface tension × surface area = 4π r² T. Ratio of work for two bubbles: W₁/W₂ = r₁² / r₂². Given radii 3:4, ratio becomes 9:16. Analogy: Stretching a balloon requires more work for larger radius due to increased surface area. Summary: work ratio depends on square of radii ratio.

Explanation: Volume is conserved. Original drop radius R, volume V = 4/3 π R³. Dividing into 8 equal droplets: volume of each v = V/8, so radius r = R / 2. Key concept: r³ scales with volume fraction. Analogy: Splitting a chocolate ball into smaller equal balls reduces size proportional to cube root of fraction. Summary: Radius of each smaller drop is half of original.

Explanation: Molecules at the surface experience NET inward cohesive forces, giving higher potential energy compared to molecules inside, which are symmetrically pulled in all directions. Key concept: Surface tension arises due to this energy difference. Analogy: Molecules at the edge of a trampoline are stretched, storing more energy than those in the center. Summary: Surface molecules have greater potential energy than interior molecules.

Explanation: Surface energy U = T × surface area. Increasing area by A increases energy by ΔU = T × A. Key concept: Surface tension measures energy per unit area. Analogy: Stretching a rubber sheet slightly increases its stored energy proportional to the stretched area. Summary: Change in surface energy is proportional to surface tension and area increase.

Explanation: The Mass of water is conserved when a drop divides; total Mass remains the same. Volume and Mass are proportional, but surface area increases due to formation of new surfaces, which requires energy. Analogy: Dividing a chocolate ball into two halves keeps total chocolate the same but increases surface exposure. Summary: Sum of masses equals original drop, though surface area changes.

Explanation: Molecules at the surface have unbalanced cohesive forces leading to higher potential energy. Kinetic energy remains similar to interior molecules at given temperature. Analogy: Edge of a trampoline is more stretched (higher potential energy) than the center. Summary: Surface molecules possess maximum potential energy compared to interior molecules.

Explanation: Surface tension T is energy per unit area required to increase surface. Thus, numerically T = surface energy / unit area. Key concept: Surface tension links Molecular forces to measurable energy. Analogy: Pulling a soap film slightly increases energy proportional to area. Summary: Surface tension equals surface energy per unit area.

Explanation: Adding Salt increases ionic interactions at the surface, leading to stronger cohesive forces and slightly higher surface tension. Key concept: Electrolytes can modify intermolecular forces. Analogy: Dissolved sugar or Salt can make a syrup layer more resistant to stretching. Summary: Salt generally increases surface tension of water.

Explanation: Sugar molecules form hydrogen bonds with water but do not significantly disrupt cohesion. Therefore, surface tension increases slightly due to stronger intermolecular forces. Analogy: Adding sugar thickens water slightly, making its surface “tighter.” Summary: Sugar increases the surface tension of water.

Explanation: Soap molecules are surfactants, reducing cohesive forces among water molecules at the surface. This lowers surface tension compared to pure water. Analogy: Soap spreads like a lubricant, allowing water surface to stretch more easily. Summary: Soap decreases the surface tension of water.

Explanation: Soap reduces water’s surface tension, allowing it to penetrate fibers and suspend dirt particles in micelles. Key concept: Reduced surface tension improves wetting and cleaning efficiency. Analogy: Soap acts like a mediator between water and dirt, helping them mix. Summary: Soap lowers surface tension, aiding in cleaning.

Explanation: Volume (and thus Mass) is conserved when a drop divides. Surface energy increases because total surface area increases. Analogy: Splitting a chocolate ball preserves total Mass but increases exposed area. Summary: Sum of volumes of smaller droplets equals original drop.

Explanation: Merging reduces total surface area, decreasing surface energy. This energy difference is released, often as Heat or kinetic energy. Key concept: Surface energy depends on area; reducing area releases energy. Analogy: Two soap bubbles combine, releasing excess energy as the surface shrinks. Summary: Energy is released during coalescence due to decreased surface area.

Explanation: This question uses Kepler’s third law: the square of orbital period T is proportional to the cube of orbital radius r, T² ∝ r³. If the radius triples, T’² / T² = (3r / r)³ = 27. Therefore, T’ = √27 × 365 ≈ 1897 days. Analogy: Larger racetrack requires more time to complete a lap at same speed. Summary: Orbital period increases with the 3/2 power of radius.

Explanation: For circular orbit, orbital speed v = √(GM / r). Reducing radius by 5%: r’ = 0.95 r, so v’ = √(GM / 0.95 r) ≈ v / √0.95 ≈ 1.025 v. Key concept: Orbital speed increases as radius decreases. Analogy: Cars on smaller tracks need higher speed to maintain the curve. Summary: Speed increases slightly (~2.5%) when orbital radius is reduced by 5%.

Explanation: Stopping distance s ∝ v² / 2a. Initial speed v₁ = 72 km/h, distance s₁ = 5 m. For v₂ = 92 km/h, ratio of distances: s₂ / s₁ = (v₂ / v₁)² ≈ (92/72)² ≈ 1.63. Therefore, s₂ ≈ 1.63 × 5 m ≈ 8.16 m. Analogy: Faster car requires longer distance to stop, like a heavier object sliding on ice. Summary: Stopping distance increases with square of speed.

Explanation: The resultant direction θ from Vector addition: tan θ = (v sin 60°) / (v + v cos 60°) = (40 × √3/2) / (40 + 40 × 0.5) = 34.64 / 60 ≈ 0.577, so θ ≈ 30°. Analogy: Two people pushing at an angle, the NET direction is between their directions. Summary: Resultant Vector points at an intermediate angle, calculated using Vector components.

Explanation: Orbital speed v = √(gR² / (R + h)) where R is Earth’s radius (~6371 km), h = 5000 km. Substituting: v ≈ √(10 × 6371² / (6371 + 5000)) km/s ≈ 6 km/s. Key concept: Centripetal acceleration for circular orbit equals gravitational acceleration. Analogy: Swinging a ball on a string; faster spin required for larger radius. Summary: Satellite speed depends on gravity and orbital radius.

Explanation: Gravitational force F = GMm / r². Surface radius R, orbit radius = 2R. force F’ = GM × 100 / (2R)² = (1/4) × (GM × 100 / R²). Surface 10 kg experiences 100 N ⇒ GM / R² = 10 N/kg. So F’ = 250 N. Analogy: Doubling distance from center reduces force by factor of 4. Summary: force decreases with square of orbital radius.

Explanation: Shortest path means boat moves perpendicular to river. Time t = 10 min = 1/6 hr. Distance across = 0.5 km ⇒ boat’s speed component across river = 0.5 / (1/6) = 3 km/h. Relative river speed along flow v_r = √(10² – 3²) = √91 ≈ 9.54 km/h ≈ 9 km/h. Analogy: Swimming across a flowing river, boat adjusts direction to minimize travel path. Summary: River’s speed can be found using Pythagoras theorem.

Explanation: Using v² = u² + 2as, u = 0, v = 500 m/s, s = 1 m ⇒ a = v² / 2s = 500² / 2 ≈ 125,000 m/s². Time t = (v – u)/a = 500 / 125,000 ≈ 0.004 s = 4 ms. Analogy: Car accelerating rapidly along a very short track reaches high speed in milliseconds. Summary: Time in barrel is very small due to high acceleration.

Explanation: Gravity g_p = GM / R². Relative to Earth: g_p / g = (1/60) / (1/2)² = (1/60) / (1/4) = 1/15. Summary: Planet’s surface gravity is 1/15 of Earth’s gravity.

Explanation: Using s = ut + ½ at², s = 25 m, u = 8 m/s, t = 8 s. 25 = 8 × 8 + ½ a × 64 ⇒ 25 = 64 + 32a ⇒ 32a = -39 ⇒ a ≈ -1.22 m/s². Analogy: Vehicle moving forward but decelerating to cover shorter distance. Summary: Acceleration can be found from displacement, initial velocity, and time.

Explanation: Scalar (dot) product: P·Q = |P||Q|cosθ. When θ = 180°, cos180° = -1, so P·Q = -|P||Q|. Key concept: Vectors pointing in exactly opposite directions are called antiparallel. Analogy: Two people pulling on a rope in opposite directions. Summary: Two Vectors at 180° are antiparallel.

Explanation: A Vector of equal magnitude pointing opposite to a given vector is called a negative vector. Key concept: Direction matters in Vectors; magnitude alone is insufficient to describe a vector fully. Analogy: Walking 5 m east vs 5 m west; same magnitude, opposite direction. Summary: Opposite direction Vectors are negative Vectors.

Explanation: Orbital speed v = √(GM / r). Given v_A = 4v, r_A = 4R, v_B = 5v. Using (v_B / v_A)² = r_A / r_B ⇒ (5/4)² = 4R / r_B ⇒ r_B = 4R × (16/25) = 2.56 R. Analogy: Faster orbit requires smaller radius for given gravitational parameter. Summary: Radius of satellite B determined from speed-radius relationship.

Explanation: NET displacement is vector sum. Horizontal: 500 – 200 = 300 m north. Vertical: 1000 m upward. Total displacement = √(300² + 1000²) ≈ √(90000 + 1000000) ≈ 1044 m. Analogy: Moving along two perpendicular axes, displacement found via Pythagoras. Summary: NET displacement calculated using vector components.

Explanation: Geostationary satellites at ~36,000 km altitude have orbital speed v = √(GM / r) ≈ 3.07 km/s. Key concept: Orbital velocity decreases with distance from Earth. Analogy: Swinging a ball on a string, larger radius requires slower spin for stability. Summary: Orbital speed depends on central mass and orbit radius.

Explanation: Maximum resultant occurs when forces act in the same direction: R_max = F₁ + F₂ = 2 + 6 = 8 N. Key concept: Vector addition along same line yields sum of magnitudes. Analogy: Two people pushing a box together in same direction move it with combined force. Summary: Maximum tension equals sum of forces along same direction.

Explanation: Escape velocity v_e = √(2GM / R). If horizontal velocity ≥ v_e, satellite overcomes gravity and leaves Earth’s orbit, moving away indefinitely. Analogy: Throwing a ball fast enough to leave a planet. Summary: Satellite escapes Earth’s gravitational influence if speed ≥ escape velocity.

Explanation: Stopping distance s ∝ v². Ratio s₁ : s₂ = (2u)² : (4u)² = 4 : 16 = 1 : 4. Analogy: Faster car needs longer braking distance. Summary: Stopping distances scale with square of speed.

Explanation: g’ above surface: g’ = g (R / (R + h))² ≈ g (6371 / 6380)² ≈ 9.772 m/s². g’ below surface: g’ = g (1 – h/R) ≈ 9 (1 – 9/6371) ≈ 9.786 m/s². Analogy: Gravity slightly weaker with altitude, slightly stronger below surface. Summary: Gravity varies with height and depth linearly/sublinearly.

Explanation: Least count (LC) = Value of one main scale division – Value of one secondary scale division. One MSD = 12 / 200 = 0.06 cm. One SSD = 6 / 20 = 0.3 cm? Check formula: LC = MSD – SSD = 0.06 – 0.03 = 0.03 cm. Key concept: LC is smallest measurable length. Analogy: Vernier calipers measure finely by subtracting scales. Summary: Least count is difference between main and vernier scale divisions.

Explanation: Gravity decreases with height: g’ = g (R / (R + h))². Setting g’ = g/3 ⇒ (R / (R + h))² = 1/3 ⇒ R + h = R√3 ⇒ h = R(√3 – 1) ≈ 0.73 R. Analogy: Gravity weakens as you move farther from Earth. Summary: Height where g = g/3 is proportional to Earth’s radius.

Explanation: Kepler’s third law: T² ∝ r³. T₁ / T₂ = √(r₁³ / r₂³) = √(4³ / 6³) = √(64 / 216) = √(0.2963) ≈ 0.54. Analogy: Outer planets take longer to orbit the Sun. Summary: Orbital periods ratio depends on cube root of orbital radii.

Explanation: Displacement x = ∫v dt. ∫₀² (4t² + 9t + 1) dt = (4/3)t³ + (9/2)t² + t |₀² = (32/3) + 18 + 2 ≈ 30.66 m. Analogy: Integrating velocity over time gives total distance covered. Summary: Displacement obtained by integrating velocity function over time.

Explanation: Escape velocity v_e = √(2gR). Ratio: v_e(Earth)/v_e(planet) = √(g_E R_E / g_P R_P) = √(4/1 × 5/1) = √20 ≈ 4.5. Analogy: Higher gravity and larger radius increases escape speed. Summary: Escape velocity ratio depends on √(g × R).

Explanation: Velocity v = ∫a dt = ∫₀⁴ 4(t – 1) dt = 4[(t²/2) – t]₀⁴ = 4[(16/2 – 4) – 0] = 4(8 – 4) = 16 m/s. Analogy: Velocity increases as area under acceleration-time graph. Summary: Integrate acceleration function over time to find velocity.

Explanation: Cross product is anti-commutative: A × B = – (B × A). Key concept: Direction of cross product is perpendicular to plane of Vectors. Analogy: Right-hand rule shows opposite directions for A × B and B × A. Summary: Assertion is true or false depending on vector properties; cross product is not commutative.

Explanation: PE = mgh ⇒ relative error ΔPE/PE = Δm/m + Δg/g + Δh/h = 1% + 2% + 4% = 7%. Analogy: Error in each measurement adds linearly for multiplication/division. Summary: Maximum error in PE equals sum of individual relative errors.

Explanation: Surface energy U = γ × surface area. Original drop radius r, final radius r’ = r / 5 (since 125 = 5³). ΔU = γ(125 × 4πr’² – 4πr²) = γ × 4π(r²(5² – 1))? Check: Energy increases because smaller drops have larger total surface area. Analogy: Breaking soap bubble into smaller bubbles increases surface area, requiring energy. Summary: Energy increases due to formation of more surface area.

Explanation: Surface energy U = 4πR² T. W₂ / W₁ = (4π(2R)² T) / (4πR² T) = 4. Summary: Doubling radius quadruples surface area, hence work done increases fourfold. Analogy: Stretching a balloon larger requires more energy proportional to area increase.

Explanation: Volume conserved: R_big³ = 1000 × r_small³ ⇒ R_big / r_small = 10. Surface energy U ∝ R² ⇒ U_big / U_small_total = (10²) / 1000 = 100 / 1000 = 1 : 10. Analogy: Merging small droplets reduces total surface area, lowering energy. Summary: Surface energy decreases as smaller droplets coalesce.

Explanation: Work done = change in surface energy = γ × ΔA. Initial area A₁ = 4πR², final A₂ = 4π(2R)² = 16πR². ΔA = 16πR² – 4πR² = 12πR². W = γ × ΔA = 30 × 12π × (2 cm)² ≈ 2261 erg. Analogy: Inflating a bubble requires energy proportional to increase in surface area. Summary: Work done is proportional to increase in surface area of the bubble.